A motorcycle and a police car are moving toward one another. The police car emits sound with a frequency of 523 Hz and has a speed of 32.2 m > s. The motorcycle has a speed of 14.8 m > s. What frequency does the motorcyclist hear?

The change in magnetic flux when the coil is rotated is equivalent to the initial flux through the coil. The magnitude of the average induced emf is found by dividing the change in magnetic flux by the time over which it occurs, multiplied by the number of turns in the coil, following Faraday's law.

The question is about applying Faraday's law of electromagnetic induction to calculate the change in magnetic flux and the induced electromotive force (emf).

Part A: Change in Magnetic Flux

The magnetic flux \\(\Phi\\) through a coil is given by \\(\Phi = B \cdot A \cdot \cos(\theta)\\), where \\(B\\) is the magnetic field strength, \\(A\\) is the area of the coil, and \\(\theta\\) is the angle between the field and the normal to the coil's plane. Initially, with the coil parallel to the field, \\(\theta = 0\\) degrees, so \\(\cos(\theta) = 1\\), and the initial flux is \\(\Phi_{initial} = B \cdot A\\). After rotation, \\(\theta = 90\\) degrees, so \\(\cos(\theta) = 0\\), and the final flux is \\(\Phi_{final} = 0\\). The change in flux is the final minus the initial. Therefore, the change in magnetic flux is equal to the initial flux \\(\Phi_{initial}\\).

Part B: Magnitude of the Average Induced Emf

To find the magnitude of the average induced emf, Faraday's law is used, which states that the induced emf is proportional to the rate of change of magnetic flux. It is given by the equation \\(emf = -N \cdot \Delta\Phi / \Delta t\\), where \\(N\\) is the number of turns in the coil, \\(\Delta\Phi\\) is the change in flux, and \\(\Delta t\\) is the time over which the change occurs. Plugging in the values given, including the rate of change of flux from part A, we can calculate \\(emf\\).

Answer:

The distance from the top of the stick would be 2l/3

Explanation:

Let the impulse 'FΔt' acts as a distance 'x' from the hinge 'H'. Assume no impulsive reaction is generated at 'H'. Let the angular velocity of the rod about 'H' just after the applied impulse be 'W'. Also consider that the center of percussion is the point on a bean attached to a pivot where a perpendicular impact will produce no reactive shock at the pivot.

Applying impulse momentum theorem for linear momentum.

FΔt = m(Wl/2), since velocity of center of mass of rod  = Wl/2

Similarly applying impulse momentum theorem per angular momentum about H

FΔt * x = I * W

Where FΔt * x represents the impulsive torque and I is the moment of inertia

F Δt.x = (ml² . W)/3

Substituting FΔt

M(Wl/2) * x = (ml². W)/3

1/x = 3/2l

x = 2l/3

so it is not bad not good. pefectly balanced just as all things should be

The radius of the orbits and the rocket's mass, The changes would involve transitioning from the elliptical to the circular orbit, taking into account Earth's gravitational pull.

This question delves into detail regarding the physics of celestial movements, in particular, satellites and their orbits. The problem pertains to the transition from an elliptical trajectory to a circular one.

The change in speed at point A would need to equal the difference between the speed in the elliptical orbit at the apogee and the speed necessary to maintain the desired circular orbit, factoring in the gravitational pull of Earth. The change in speed at point B would need to address any residual speed from the elliptical orbit to ensure the rocket's speed aligns with the necessary speed to maintain the circular orbit around the Earth.

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Answer:

5.4 rad/s

48.6 m/s

5.4 m/s²

262.44 m/s²

Explanation:

r = Radius of centrifuge = 9 m

[tex]\theta=0.3t^2[/tex]

Differentiating with respect to time

[tex]\frac{d\theta}{dt}=0.6t=\omega[/tex]

At t = 9 s

[tex]\omega=0.6\times 9=5.4\ rad/s[/tex]

Angular velocity is 5.4 rad/s

Linear speed

[tex]v=r\omega\\\Rightarrow v=9\times 5.4=48.6\ m/s[/tex]

The linear speed of the astronaut is 48.6 m/s

Differentiating [tex]\omega[/tex] with respect to time

[tex]\frac{d\omega}{dt}=0.6=\alpha[/tex]

Tangential acceleration is given by

[tex]a_t=\alpha r\\\Rightarrow a_t=0.6\times 9=5.4\ m/s^2[/tex]

Tangential acceleration of the astronaut is 5.4 m/s²

Radial acceleration is given by

[tex]a_r=\frac{v^2}{r}\\\Rightarrow a_r=\frac{48.6^2}{9}\\\Rightarrow a_r=262.44\ m/s^2[/tex]

The radial acceleration of the astronaut is 262.44 m/s²

To solve this problem it is necessary to resort to concepts related to the kinematic equations of movement description which define speed as

[tex]v = \frac{x}{t}[/tex]

Where,

x = Distance

t=Time

Re-arrange to find the time we have,

[tex]t = \frac{x}{v}[/tex]

At 20°C the speed of sound in air is 343.2m/s and the speed of sound in water is 1484.3m/s

The time in which the sound lasts in traveling in the air would be given by the shipping and return (even when it touches the water) which is 28m (14 way and 14 back), in other words

[tex]t_a = \frac{28}{343.2} = 0.08158s[/tex]

In the case of the lake, the same logic is followed, only that there is no net distance here, therefore

[tex]t_w = \frac{2d}{1484.3}[/tex]

The total time would be given by

[tex]t_T = 0.08158 + \frac{2d}{1484.3}[/tex]

The total time is given in the statement so we clear to find d

[tex]0.108-0.08158= \frac{2d}{1484.3}[/tex]

[tex]d = \frac{1484.3(0.108-0.08158)}{2}[/tex]

[tex]d = 18.86m[/tex]

Therefore the lake has a depth of 18.86 m

Answer:

Centripetal force, [tex]F_x=0.379\ N[/tex]

Explanation:

It is given that,

Mass of the ball, m = 62 g = 0.062 kg

Length of the string, l = 72 cm = 0.72 m

Angle, [tex]\theta=32^{\circ}[/tex]

To find,

The centripetal force experienced by the ball.

Solution,

According to the attached free body diagram,

[tex]T\ cos\theta-mg=0[/tex]

[tex]T=\dfrac{mg}{cos\theta}[/tex]

[tex]T=\dfrac{0.062\times 9.8}{cos(32)}[/tex]

T = 0.716 N

The centripetal force will be experienced by the ball in horizontal direction. It is equal to :

[tex]F_x=T\ sin\theta[/tex]

[tex]F_x=0.716\ N\ sin(32)[/tex]

[tex]F_x=0.379\ N[/tex]

So, the centripetal force experienced by the ball is 0.379 N.

The centripetal force experienced by the small ball if mass 62 g and attached to the string is 0.379 N.

Centripetal force is the force which is required to keep rotate a body in a circular path. The direction of the centripetal force is inward of the circle towards the center of rotational path.

Given information-

The mass of the small bag is 62 grams.

The length of the string is 72 cm.

Suppose T is the tension in the string, m is the mass of ball and g is the gravitational force.

The horizontal force applied on the body is,

[tex]\sum F_x=T\cos \theta-mg=0\\T=\dfrac{mg}{\cos \theta}[/tex]

The centripetal force applied on the body is,

[tex]F_c=T\sin \theta\\T=\dfrac{F_c}{\sin \theta}[/tex]

Compare both the values of the T, we get,

[tex]\dfrac{F_c}{\sin\theta}=\dfrac{mg}{\cos \theta}\\{F_c}=\dfrac{mg\times{\sin\theta}}{\cos \theta}\\{F_c}={mg\times{\tan\theta}}[/tex]

Put the values as,

[tex]F_c=0.062\times9.8\times\tan(32)\\F_c=0.379\rm N[/tex]

Thus, the centripetal force experienced by the small ball if mass 62 g and attached to the string is 0.379 N.

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Answer:

77.83783 °C

Explanation:

V = Volume of room = 36 m²

c = Heat capacity = 1000 J/kg C

P = Power = 120 J/s = 120 W

t = Time = 8 hours

[tex]\rho[/tex] = Density of air = 1.225 kg/m³

[tex]\Delta T[/tex] = Change in temperature

Density is given by

[tex]\rho=\frac{m}{V}\\\Rightarrow m=\rhoV\\\Rightarrow m=1.225\times 36\\\Rightarrow m=44.1\ kg[/tex]

Heat is given by

[tex]Q=mc\Delta T\\\Rightarrow Pt=mc\Delta T\\\Rightarrow \Delta T=\frac{Pt}{mc}\\\Rightarrow \Delta T=\frac{120\times 8\times 3600}{44.4\times 1000}\\\Rightarrow \Delta T=77.83783^{\circ}C[/tex]

The temperature change you expected in this air 77.83783 °C

Final answer:

The estimated temperature change in the air of your bedroom while you sleep is approximately 78.3 °C.

Explanation:

In order to estimate the temperature change in the air of your bedroom while you sleep, we need to calculate the amount of thermal energy emitted by your body and the amount of energy absorbed by the air. Your metabolic rate is given as P = 120 J/s. Assuming you sleep for 8.0 hours, the total energy emitted is 120 J/s * 8 hours * 3600 seconds/hour = 3456000 J.

The specific heat capacity of air is given as 1000 J/kg C and the volume of your room is 36 m³. Assuming the density of air is 1.225 kg/m³, the mass of the air in your room is 1.225 kg/m³ * 36 m³ = 44.1 kg.

Using the equation Q = mcΔθ, where Q is the thermal energy absorbed/lost, m is the mass, c is the specific heat capacity, and Δθ is the change in temperature, we can calculate the temperature change as follows:

Q = mcΔθ

Δθ = Q / mc

Δθ = 3456000 J / (44.1 kg * 1000 J/kg C)

Δθ ≈ 78.3 °C

Therefore, we can estimate that the temperature change in the air of your bedroom is approximately 78.3 °C.

The speed of the car is 60 m/s

Explanation:

The speed of an object is a scalar quantity telling "how fast" the object is moving, regardless of its direction. It can be calculated as follows:

[tex]speed=\frac{d}{t}[/tex]

where

d is the distance covered

t is the time taken

For the car in this problem, we have

[tex]d=108 km =1.08\cdot 10^5 m[/tex] is the distance covered

[tex]t=30 min \cdot 60 =1800 s[/tex] is the time taken

Subsituting into the equation, we find the speed:

[tex]speed=\frac{1.08\cdot 10^5}{1800}=60 m/s[/tex]

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Answer:

[tex]T = m*g*cos\theta[/tex]

Explanation:

Since tarzan moves in a circular trajectory we can sum all forces on the centripetal-axis:

[tex]T - m*g*cos\theta = m*a_c[/tex]

[tex]T - m*g*cos\theta = m*V^2/R[/tex]   Since the speed is zero:

[tex]T - m*g*cos\theta = 0[/tex]

[tex]T = m*g*cos\theta[/tex] Having Tarzan's mass we could calculate the module of the tension in the vine.

The tension in Tarzan's vine can be determined at the highest point of his swing, when his velocity is zero, using Newton's laws of motion. At this point, the tension's vertical component equals his weight, while the horizontal component, providing centripetal force, is zero. The equation T=mg/cosθ can be used to calculate the tension.

The subject of your question involves physics, particularly mechanics. The tension in the vine at the highest point of Tarzan's swing, when his velocity is zero and the vine makes an angle of 40 degrees with the vertical, can be found using Newton's second law of motion, which can be written as ΣF=ma, where F represents force, m is mass, and a is acceleration.

At the highest point of the swing, the only forces acting on Tarzan are the tension in the vine (T) and his weight (mg), where m is his mass and g is the acceleration due to gravity. These two forces result in a net force that provides the centripetal acceleration necessary for Tarzan to change direction and swing back down. We have:

So we can get the tension in vine (T) from the equation Tcosθ=mg. Rearranging, we have T=mg/cosθ.

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The orbital period of a spacecraft can be calculated using Kepler's Third Law and the kinetic energy can be determined using the equation K = GMm/(2r).

The orbital period of the spacecraft can be found using Kepler's Third Law, which states that the square of the period of a body moving in a circular orbit is proportional to the cube of the average distance from the center of the force. If we let RE and m denote Earth's radius (6.371 x 10^6 m) and mass (5.972 x 10^24 kg), and G define the gravitational constant (6.67 x 10^-11 N m^2/kg^2), the period T (in seconds) can be calculated as T = 2π √[(4π^2(R+2h)^3)/Gm], or T = 2π √[(4π^2(3RE)^3)/(GmE)].

The kinetic energy K of an object moving in a circular orbit is given by K = GMm/(2r), where M is the mass of the central body (in this case Earth), m is the mass of the object (the spacecraft), and r is the distance of the object from the center of the force (in this case the Earth's center). In this instance, K = GMm/(2R), with R being the radius of Earth, should suffice.

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Answer:

D₂ = 2.738 cm

Explanation:

Continuity equation

The continuity equation is nothing more than a particular case of the principle of conservation of mass. It is based on the flow rate (Q) of the fluid must remain constant throughout the entire pipeline.

Since the flow rate is the product of the surface of a section of the duct because of the speed with which the fluid flows, we will have to comply with two points of the same pipeline:  

Q = v*A  : Flow Equation

where:

Q = Flow in (m³/s)

A is the surface of the cross sections of points 1 and 2 of the duct.

v is the flow velocity at points 1 and 2 of the pipe.

It can be concluded that since the flow rate must be kept constant throughout the entire duct, when the section decreases, the flow rate increases in the same proportion and vice versa.

Data

D₁= 6.0 cm : faucet  diameter

v₁ = 5 m/s :  speed of fluid in the  faucet

v₂ = 20 m/s : speed of fluid in the  nozzle

Area calculation

A = (π*D²)/4

A₁ = (π*D₁²)/4

A₂ = (π*D₂²)/4

Continuity equation  

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁(π*D₁²)/4 = v₂(π*D₂²)/4 : We divide by (π/4) both sides of the equation

v₁ (D₁)² = v₂(D₂)²

We replace data

6 *(5)² = 20*(D₂)²

150 = 20*(D₂)²

(150 /20) = (D₂)²

7.5 = (D₂)²

[tex]D_{2} = \sqrt{7.5}[/tex]

D₂ = 2.738 cm :  nozzle diameter

Answer:

T = 451.26 N

Explanation:

It is given that,

The mass of block, m = 46 kg

Mass of the chain, m' = 19 kg

Length of the chain, l = 1.9 m

Let T is the the tension in the chain at the point where the chain is supporting the block. It is clearly equal to the product of mass and acceleration.

[tex]T=mg[/tex]

[tex]T=46\ kg\times 9.81\ m/s^2[/tex]

T = 451.26 N

So, the tension in the chain at the point where the chain is supporting the block is 451.26 N. Hence, this is the required solution.

Answer:

496.57492 kg/m³

Explanation:

[tex]P_a[/tex] = Atmospheric pressure = 101300 Pa

[tex]\rho_w[/tex] = Density of water = [tex]1000 kg/m^3[/tex]

[tex]h_w[/tex] = Height of water = 0.221 m

[tex]h_l[/tex] = Height of fluid = 0.335 m

g = Acceleration due to gravity = 9.81 m/s²

[tex]\rho_l[/tex] = Density of the unknown fluid

Absolute pressure at the bottom

[tex]P_{abs}=P_a+\rho_wgh_w+\rho_lgh_l\\\Rightarrow \rho_l=\frac{P_{abs}-P_a-\rho_wgh_w}{gh_l}\\\Rightarrow \rho_l=\frac{104900-101300-1000\times 9.81\times 0.221}{9.81\times 0.335}\\\Rightarrow \rho_l=435.73873\ kg/m^3[/tex]

The density of the unknown fluid is 496.57492 kg/m³

To find the density of the unknown fluid in the container, we subtract the atmospheric pressure and the pressure due to water from the absolute pressure at the bottom of the container and divide the result by the product of gravity and the height of the unknown fluid.

The question is asking about the determination of the density of an unknown fluid in an open container where you have two separated fluids due to their different densities. In this setup, water fills the lower portion of the container and the unknown fluid fills the upper part.

The pressure at the bottom of this container (P) can be expressed as the sum of the atmospheric pressure (P0), pressure due to water (ρw * g * h1), and pressure due to the unknown fluid (ρl * g * h2) where 'g' is the acceleration due to gravity and 'h' represents the depth of each fluid. Given in the problem are P, P0, ρw, g, h1, and h2 and to find the density of the unknown fluid (ρl), we rearrange the equation into ρl = (P - P0 - ρw * g * h1) / (g * h2). By substituting the given values into this equation, we can find the density of the unknown fluid.

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Answer:

The amplitude is 0.010 m

Solution:

As per the question:

Eccentric Mass, [tex]m_{e} = 10\ kg[/tex]

[tex]m_{e}[/tex] = 10%M kg

0.1M = 10

Total mass, M = 100 kg

Spring constant, k = 3200 N/m

Eccentric center, e = 100 mm = 0.1 m

Speed of motor, N = 1750 rpm

Distance between two springs, d = 250 mm = 0.25 m

Now,

Angular velocity, [tex]\omega = \frac{2\pi N}{60} = \frac{2\pi \times 1750}{60} = 183.26\ rad/s[/tex]

For the vertical vibrations:

[tex]\omega_{n} = \sqrt{\frac{2k}{M}} = \sqrt{\frac{6400}{100}} = 8\ rad/s[/tex]

Now, the frequency ratio is given by:

r = [tex]\frac{\omega}{\omega_{n}} = \frac{183.26}{8} = 22.91[/tex]

Thus the amplitude is given by:

A = [tex]e^{\frac{m_{e}}{M}}.\frac{r^{2}}{|1 - r^{2}|}[/tex]

A = [tex]e^{\frac{10}{100}}.\frac{22.91^{2}}{|1 - 22.91^{2}|} = 0.010 m[/tex]

Final answer:

The amplitude of vertical vibration for the electric motor mounted on springs is 100 mm or 0.1 meters, considering the eccentric mass and neglecting damping.

Explanation:

The subject of this question is Physics, and it pertains to a college-level study of mechanics in the context of oscillations and waves, specifically relating to a system experiencing simple harmonic motion (SHM). The motor with an eccentric mass which causes vertical vibrations can be modeled as a mass-spring system, characteristic of simple harmonic oscillators.

Given the mass and the spring constant, we can calculate the natural angular frequency (not represented in this problem due to lack of damping) of this system. The driving frequency can also be converted to an angular frequency to check against the natural frequency. However, the question directly asks for the amplitude of vertical vibration, which in the undamped case of SHM, depends on the initial displacement of the mass from its equilibrium position. Since the question neglects damping, the amplitude can be considered equal to the eccentricity of the mass in the static case. Therefore, the amplitude of vertical vibration is 100 mm, or 0.1 meters.

Answer:

2.79 m

Explanation:

Use the static equilibrium condition, net torque actin on the system is zero.

[tex]\sum \tau= 0[/tex]

[tex]T(Lsin\theta)- Mg\frac{L}{2}-mgx=0[/tex]

solve for the distance of the person from the wall x

[tex]x= \frac{TLsin\theta-Mg(L/2)}{Mg}[/tex]

now putting the values we get

[tex]x= \frac{1350\times4.25sin30-47\times9.81\times(4.25/2)}{69\times9.80}[/tex]

= 2.79 m

Answer:

Answered

Explanation:

a) The magnitude of the magnetic flux through the coil before it is rotated is,

[tex]\phi_{initial}= BAsin\theta[/tex]

=BAsin90° = BA

therefore, the magnitude of the magnetic flux= BA

b) The magnitude of the magnetic flux through the coil after it is rotated is, Ф_final = BAsinθ

= BA sin0= 0

therefore, the magnitude of magnetic flux through the coil after it is rotated is, zero.

c)  The magnitude of the average emf induced in the coil is,

[tex]\epsilon = N\frac{d\phi}{dt}[/tex]

[tex]\epsilon = N\frac{NBA}{\Delta t}[/tex]

because Ф = BA

Answer:

The tension in the rope is 20 N

Solution:

As per the question:

Mass of the object, M = 2 kg

Density of water, [tex]\rho_{w} = 1000\ kg/m^{3}[/tex]

Density of the object, [tex]\rho_{ob} = 500\kg/m^{3}[/tex]

Acceleration due to gravity, g = [tex]10\ m/s^{2}[/tex]

Now,

From the fig.1:

'N' represents the Bouyant force and T represents tension in the rope.

Suppose, the volume of the block be V:

V = [tex]\frac{M}{\rho_{ob}}[/tex]              (1)

Also, we know that Bouyant force is given by:

[tex]N = \rho_{w}Vg[/tex]

Using eqn (1):

[tex]N = \rho_{w}\frac{M}{\rho_{ob}}g[/tex]

[tex]N = 1000\frac{2}{500}\times 10 = 40\ N[/tex]

From the fig.1:

N = Mg + T

40 = 2(10) + T

T = 40 - 20 = 20 N

[tex]N = \rho_{w}Vg[/tex]

The question involves using the Conservation of Energy to find the speed of an object in simple harmonic motion. The concept relies on the fact that the total energy in the harmonic system remains constant, allowing us to determine the kinetic energy of the object and consequently its speed.

This question can be answered using the Conservation of Energy principle.

In the given problem, the object attached to the spring performs simple harmonic motion, meaning that the total energy of the system (kinetic energy + potential energy) remains constant.

At any instant in time, the total mechanical energy of the object will be:

E = 1/2 k A²

Where.

E is total energy, k is the spring constant, and A is the system's amplitude.

When the displacement of the object is at x = 0.040 m, the potential energy of the object (U) will be:

U = 1/2  k x²

At this point, the kinetic energy (K) can be found by subtracting the potential energy (U) from the total energy (E).

K = E - U

Knowing that the kinetic energy (K) is also given by 1/2 m  v² (where m is mass and v is speed), we can rearrange this equation to find the speed of the object:

v = sqrt((2K)/m)

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The speed of the object at a displacement of 0.040 m is 0.49 m/s.

Calculating Speed in Simple Harmonic Motion Using Conservation of Energy

To find the speed of a 0.20 kg object attached to a spring (k = 10 N/m) at a displacement of 0.040 m, we can use the principle of conservation of energy.

Calculate the total mechanical energy (E) in the system using the amplitude (A):

E = 1/2 x k x [tex]A^2[/tex]

E = 1/2 x 10 N/m x [tex](0.080 m)^2[/tex]

E = 0.032 J

Determine the potential energy (U) at the displacement (x = 0.040 m):

U = [tex]1/2 \times k \times x^2[/tex]

U = [tex]1/2 \times 10 N/m \times (0.040 m)^2[/tex]

U = 0.008 J

Find the kinetic energy (K) at that displacement:

K = E - U

K = 0.032 J - 0.008 J

K = 0.024 J

Calculate the speed (v) using the kinetic energy:

K = 1/2 x m x [tex]v^2[/tex]

[tex]0.024 J = 1/2 \times 0.20 kg \times v^2[/tex]

[tex]v^2 = 0.24 m^2/s^2[/tex]

v = 0.49 m/s

Therefore, the speed of the object at a displacement of 0.040 m is 0.49 m/s.

Nodes and antinodes on a vibrating string standing wave has different amplitudes.

Option B

Explanation:

Final answer:

Points on a tightened string vibrating with a standing wave have different amplitudes, with nodes having no displacement and antinodes having the maximum. The frequency and speed of the wave are constant along the string, but energy varies with amplitude.

Explanation:

When a tightened string vibrates with a standing wave, points on the string vibrate with different amplitudes. This variation in amplitude can be seen by observing the points of maximum displacement (antinodes) and the points of no displacement (nodes). While points at the antinodes experience the greatest amplitude, the points at the nodes do not move at all. The frequency of oscillation is the same for all points on the string because a standing wave is established by the interference of waves traveling in opposite directions with the same frequency. Consequently, the speed of the wave along the string is also the same because it is determined by the tension and the linear mass density of the string which we assume are constant. However, the energy of the points is not the same due to the varying amplitudes; points at the antinodes have more energy as compared to those at the nodes.

Answer:

v₀ = 13.9 10³ m / s

Explanation:

Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.

    F = m a

    G m M / x² = m dv / dt = m dv/dx  dx/dt

    G M / x² = dv/dx   v

    GM dx / x² = v dv

We integrate

    v² / 2 = GM (-1 / x)

We evaluate between the lower limits where x = Re = 6.37 10⁶m  and the velocity v = vo and the upper limit x = 2.50 10⁸m  with a velocity of v = 8.50 10³ m/s

    ½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))

    72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)

    72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3  10⁻⁸)

    72.25 10⁶ - v₀² = -1.213  10⁸

    v₀² = 72.25 10⁶ + 1,213 10⁸

    v₀² = 193.6 10⁶

    v₀ = 13.9 10³ m / s

Answer:

V = 8.26 m/s

Explanation:

The sum of forces on the centripetal-axis is:

[tex]Ff = m*a_c[/tex]

Maximum speed happens when friction force is maximum, so:

[tex]\mu*N = m*V^2/R[/tex]

[tex]\mu*m*g = m*V^2/R[/tex]

[tex]V=\sqrt{R*\mu*g}[/tex]

V = 8.26 m/s

The maximum speed that he can travel through the arc without slipping is mathematically given as

v = 9.93m/s

Question Parameters:

Generally the equation for the Maximum velocity   is mathematically given as

[tex]\frac{mv^2}{r} = \mu_s mg[/tex]

Therefore

[tex]v = \sqrt{(0.63)(9.8)(16)}[/tex]

v = 9.93m/s

Therefore, the maximum speed that he can travel through the arc without slipping is

v = 9.93m/s

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Answer:

a) J = 13067 kg*m/s

b) J = 9240 kg*m/s

c) F = 2666.73 N

d) F = 21488.37 N

e) 135°

Explanation:

We know that the impulse could be calculated by:

J = ΔP

where ΔP is the change in the linear momentum:

ΔP = [tex]P_f -P_i[/tex]

Also:

P = MV

Where M is the mass and V is the velocity.

so:

J= [tex]MV_f-MV_i[/tex]

where [tex]V_f[/tex] is the final velocity and [tex]V_i[/tex] is the inicial velocity

a)The impluse from turn is:

[tex]J_x=MV_{fx}-MV_{ix}[/tex]

[tex]J_y=MV_{fy}-MV_{iy}[/tex]

On the turn, [tex]V_{ix}=0[/tex] and [tex]V_{fy}=0[/tex], the magnitude of the impulse on direction x and y are:

[tex]J_x=9240 kg*m/s[/tex]

[tex]J_y=-9240 kg*m/s[/tex]

So, using pythagoras theorem the magnitude of the impulse is:

J = [tex]\sqrt{9240^2+9240^2}[/tex]

J = 13067 kg*m/s

b) The impluse from the collision is:

[tex]J=MV_{f}-MV_{i}[/tex]

J = 0 - (1400)(6.6)

J = 9240 kg*m/s

c) Using the next equation:

FΔt = J

where F is the force, Δt is the time and J is the impulse.

Replacing J by the impulse due to the turn, Δt by 4.9s and solving for F we have that:

F = J / Δt

F = 13067 / 4.9 s

F = 2666.73 N

d) At the same way, replacing J by the impulse during the collision, Δt by 0.43s and solving for F we have that:

F = J / Δt

F = 9240 / 0.43

F = 21488.37 N

e) The force have the same direction than the impulse due to the turn, Then, if the impulse have a direction of -45°, the force have -45° or 135°

To estimate the speed of the swinging orangutan, we can use the principles of rotational motion and trigonometry. By considering the angle of swing and the length of the orangutan's arm, we can calculate the horizontal distance traveled in one swing. However, without the time taken for one swing, we cannot provide an exact value for the speed.

To estimate the speed at which the orangutan is moving while swinging from branch to branch, we can use the principles of rotational motion and trigonometry. When the orangutan swings to a 20° angle, it forms a right triangle with the vertical height being 0.90 m and the hypotenuse being the length of the arm. By using the sine function, we can find the horizontal distance traveled by the orangutan in one swing.

Using the formula sin(20°) = horizontal distance / 0.90 m, we can rearrange the equation to find the horizontal distance traveled:

horizontal distance = 0.90 m * sin(20°)

Since the orangutan takes one swing immediately after another, the time taken for one swing is negligible. Therefore, the speed at which the orangutan is moving can be calculated by dividing the horizontal distance by the time taken for one swing. As the time taken is not given in the question, we cannot provide an exact value for the speed.

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To estimate the speed of an orangutan swinging, we use the physics formula for the speed of an object in pendulum motion. Under the given conditions of a 20° swing and arm length of 0.9 m, the orangutan's speed is approximately 2.54 m/s.

The behavior you're describing, where an orangutan swings like a pendulum beneath handholds, involves the study of pendulum motion, which is a concept in physics. Under simplified physics assumptions, we can model the swinging motion using the formula for the speed (v) of an object in pendulum motion: v = √(2*g*L(1-cos(θ))), where g is the acceleration due to gravity (9.8 m/s^2), L is the length of the pendulum (which we can approximate as the length of an orangutan's arm, 0.9 m), and θ is the angle swung through (20° or 0.349 radians after conversion).

Plugging the given values into the formula gives us: v = √(2*9.8*0.9(1-cos(0.349))), so the speed of an orangutan swinging would be approximately 2.54 m/s.

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Answer:

ωf = 4.53 rad/s

Explanation:

By conservation of the angular momentum:

Ib*ωb = (Ib + Ic)*ωf

Where

Ib is the inertia of the ball

ωb is the initial angular velocity of the ball

Ic is the inertia of the catcher

ωf is the final angular velocity of the system

We need to calculate first Ib, Ic, ωb:

[tex]Ib = mb*(L/2)^2=0.15*(1.2/2)^2=0.054 kg.m^2[/tex]

[tex]Ic = mc/12*L^2=2.2/12*1.2^2=0.264 kg.m^2[/tex]

ωb = Vb / (L/2) = 16 / (1.2/2) = 26.67 m/s

Now, ωf will be:

[tex]\omega f = \frac{Ib*\omega b}{Ib + Ic}  = 4.53rad/s[/tex]

Answer:

 W = 1.8 J

Explanation:

given,

mass of square plate = 0.6 Kg

length of side of square plate = 0.15 m

speed is change from   0 to 40.0 rad/s

work done = ?

moment of inertia through the plane perpendicular to plane

[tex]I = \dfrac{Mr^2}{6}[/tex]

r will be equal to 0.15 m

[tex]I = \dfrac{0.6 \times 0.15^2}{6}[/tex]

[tex]I =0.00225\ kgm^2[/tex]

work done is equal to change in kinetic energy

 [tex]W = \dfrac{1}{2}I(\omega_f^2-\omega_i^2)[/tex]

 [tex]W = \dfrac{1}{2}\times 0.00225(40^2-0^2)[/tex]

   W = 1.8 J

work done when speed changes from 0 to 40 rad/s is equal to  W = 1.8 J

Answer:

Explanation:

F(r) = Kr/|r|3

= k / r²

Work done

= ∫F(r) dr

= ∫ k / r²  dr

limit from

r₁ = 2 to r₂ = √ ( 2² + 4² + 5² ) = 6.7

Work done

∫ k / r² dr

[ - k / r ]

Taking limit from 2 to 6.7

Work done  = K ( - 1 / 6.7 + 1 / 2 )

= 0.35 K

The work done as a particle moves along a straight line from one point to another in the presence of an electric field is 0.

To find the work done as a particle moves along a straight line from one point to another in the presence of an electric field, you can use the following formula for the work done:

[tex]\[W = \int_{\mathbf{r_1}}^{\mathbf{r_2}} \mathbf{F} \cdot d\mathbf{r}\][/tex]

Where:

- [tex]\(W\)[/tex] is the work done.

- [tex]\(\mathbf{F}\)[/tex] is the force acting on the particle.

- [tex]\(d\mathbf{r}\)[/tex] is the infinitesimal displacement vector along the path of the particle.

- [tex]\(\mathbf{r_1}\)[/tex] and [tex]\(\mathbf{r_2}\)[/tex] are the initial and final positions of the particle.

In this case, the force [tex]\(\mathbf{F}\)[/tex] is given as [tex]\(F(\mathbf{r})[/tex] = [tex]\frac{K\mathbf{r}}{|\mathbf{r}|^3}\)[/tex], and the particle moves along a straight line from [tex]\(\mathbf{r_1} = (2, 0, 0)\)[/tex] to [tex]\(\mathbf{r_2} = (2, 4, 5)[/tex].

We can rewrite [tex]\(\mathbf{F}[/tex] as a vector:

[tex]\[\mathbf{F} = \frac{K}{|\mathbf{r}|^3}\mathbf{r}\][/tex]

Now, let's calculate the differential displacement [tex]\(d\mathbf{r}\)[/tex] along the path from [tex]\(\mathbf{r_1}\)[/tex] to [tex]r_2[/tex]:

[tex]\[d\mathbf{r} = (dx, dy, dz)\][/tex]

Now, let's calculate the dot product [tex]\(\mathbf{F} \cdot d\mathbf{r}\)[/tex]:

[tex]\[\mathbf{F} \cdot d\mathbf{r} = \frac{K}{|\mathbf{r}|^3}\mathbf{r} \cdot (dx, dy, dz)\][/tex]

Since the path is a straight line, [tex]\(\mathbf{r}\) and \(d\mathbf{r}\)\\[/tex] are parallel, so[tex]\(\mathbf{r} \cdot d\mathbf{r} = |\mathbf{r}||d\mathbf{r}|\cos(0^\circ) = |\mathbf{r}||d\mathbf{r}|\)[/tex].

Therefore, we have:

[tex]\[\mathbf{F} \cdot d\mathbf{r} = \frac{K}{|\mathbf{r}|^3}|\mathbf{r}||d\mathbf{r}| = \frac{K}{|\mathbf{r}|^2}|d\mathbf{r}|\][/tex]

Now, we can integrate [tex]\(\mathbf{F} \cdot d\mathbf{r}\)[/tex] from [tex]\(\mathbf{r_1}\)[/tex] to [tex]\(\mathbf{r_2}\)[/tex] along the straight line path:

[tex]\[W = \int_{\mathbf{r_1}}^{\mathbf{r_2}} \frac{K}{|\mathbf{r}|^2}|d\mathbf{r}|\][/tex]

The integration limits and the path are straight, so this simplifies to:

[tex]\[W = K \int_{2}^{2} \frac{1}{(2^2+4^2+5^2)^{3/2}}|d\mathbf{r}|\][/tex]

Since the particle moves from [tex]\(\mathbf{r_1}\) to \(\mathbf{r_2}\)[/tex], the displacement [tex]\(|d\mathbf{r}| = |\mathbf{r_2} - \mathbf{r_1}| = |(0, 4, 5)| = \sqrt{0^2 + 4^2 + 5^2} = \sqrt{41}\)[/tex].

So, the work done is:

[tex]\[W = K \int_{2}^{2} \frac{1}{(2^2+4^2+5^2)^{3/2}}\sqrt{41} dr\][/tex]

Since the limits of integration are the same, the integral evaluates to zero. Therefore, the work done as the particle moves along a straight line from (2, 0, 0) to (2, 4, 5) is zero.

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To find the translational speed of the bowling ball at the top of the vertical rise, we can use the principle of conservation of mechanical energy.

To find the translational speed of the bowling ball at the top of the vertical rise, we can use the principle of conservation of mechanical energy. Initially, the ball has kinetic energy due to its translational speed. As it moves up the vertical rise, its gravitational potential energy increases. Since there is no friction, we can assume there is no loss of energy.

Using the equation for conservation of mechanical energy:

Initial kinetic energy = final potential energy

0.5 * m * v^2 = m * g * h

Where m is the mass of the ball, v is the translational speed at the bottom, g is the acceleration due to gravity, and h is the height of the vertical rise.

Plugging in the given values:

0.5 * m * (8.57 m/s)^2 = m * 9.8 m/s^2 * 0.760 m

Simplifying and solving for the mass:

m = (9.8 m/s^2 * 0.760 m) / (0.5 * (8.57 m/s)^2)

Finally, we can plug the mass back into the initial equation to find the translational speed at the top of the rise.

Answer:

h = 15.34 m

Explanation:

given,

tuning fork vibration = 513 Hz

speed of sound = 343 m/s

frequency after deflection = 489 Hz

the source (the fork) moves away from the observer, its speed increases and hence the apparent frequency decreases

[tex]f_{apparent} = \dfrac{v}{v+u}f_0[/tex]

[tex]489 = \dfrac{343}{343+u}\times 513[/tex]

[tex]0.953 = \dfrac{343}{343+u}[/tex]

[tex]343+u = \dfrac{343}{0.953}[/tex]

[tex]343+u = 359.92[/tex]

u = 16.92 m/s

height of the building

v² = u² + 2 g s

16.92² = 2 x 9.8 x h

h = 14.61 m

time taken by sound to reach observer

[tex]t = \dfrac{14.61}{343}[/tex]

[tex]t =0.0426\ s[/tex]

in this time tuning fork has fallen one more now,

[tex]h' = u t + \dfrac{1}{2}gt^2[/tex]

[tex]h' = 16.92\times 0.0426 + \dfrac{1}{2}\times 9.8 \times 0.0426^2[/tex]

h' = 0.7296 m = 0.73 m

total distance

      h = 14.61 + 0.73

      h = 15.34 m

Final answer:

The tuning fork has fallen through a distance of 2,933.77 m.

Explanation:

To calculate the distance fallen by the vibrating tuning fork, we can use the equation:

f' = f(v/(v + vo))

where f' is the observed frequency, f is the original frequency of the tuning fork, v is the speed of sound, and vo is the velocity of the tuning fork as it falls.

Plugging in the given values: f = 513 Hz, v = 343 m/s, and f' = 489 Hz, we can rearrange the equation to solve for vo:

vo = v(f/f' - 1)

Substituting the values and solving for vo:

vo = 343((513/489) - 1) = 12.3 m/s

Since the equation for distance fallen is:

d = vo*t + (1/2)*g*t^2

and we are assuming the tuning fork is dropped from rest, we can simplify the equation to:

d = (1/2)*g*t^2

where g is the acceleration due to gravity and t is the time the tuning fork has been falling. Since we are solving for distance, we can rearrange the equation to solve for t:

t = sqrt(2*d/g)

Substituting the values and solving for t:

t = sqrt(2*d/9.8) = sqrt(2*d/10) = sqrt(d/5)

Now, we can substitute the value of vo and the known value of f' = 489 Hz into the equation:

vo*t = sqrt(d/5)*12.3 = d

Plugging in the known value of f' = 489 Hz and solving for d:

sqrt(d/5)*12.3 = d/489

Squaring both sides of the equation and solving for d:

d = (12.3^2 * 489^2)/(489^2 - 12.3^2*5) = 2933.77 m

To solve this problem it is necessary to apply the concept of period in a pendulum. Its definition is framed in the equation

[tex]T= 2\pi \sqrt{\frac{l}{g}}[/tex]

Where,

l = Length

g = Gravity

As you can see, the Period is dependent on both the gravitational acceleration and the length of the string. The period does not depend on mass. Therefore the period must remain the same at 2 seconds.

Answer:

Viscosity of blood will be [tex]918.54\times 10^{-6}Pa-sec[/tex]

Explanation:

We have given time to pass the capillary = 1.65 sec

Length of capillary L = 2 mm = 0.002 m

Pressure drop [tex]\Delta P=2.85kPa=2.85\times 10^3Pa[/tex]

Diameter [tex]d=5\mu m=5\times 10^{-6}m[/tex]

So radius [tex]r=\frac{5\times 10^{-6}}{2}=2.5\times 10^{-6}m[/tex]

Velocity will be [tex]v=\frac{L}{t}=\frac{0.002}{1.65}=1.212\times 10^{-3}m/sec[/tex]

We know that viscosity is given by [tex]\eta =\frac{r^2\Delta P}{8Lv}=\frac{(2.5\times 10^{-6})^2\times 2.85\times 10^3}{8\times 0.002\times 1.212\times 10^{-3}}=918.54\times 10^{-6}Pa-sec[/tex]

Viscosity of blood will be [tex]918.54\times 10^{-6}Pa-sec[/tex]

The viscosity of the blood flowing in the given capillary is [tex]9.27 \times 10^{-4} \ Pa.s[/tex]

The given parameters;

The velocity of the blood is calculated as follows;

[tex]v = \frac{L}{t} = \frac{2\times 10^{-3} \ m}{1.65 \ s} = 0.0012 \ m/s[/tex]

The radius of the capillary is calculated as follows;

[tex]r = \frac{d}{2} \\\\r = \frac{5\times 10^{-6}}{2} = 2.5 \times 10^{-6} \ m[/tex]

Assuming laminar flow, the viscosity of the blood is calculated as;

[tex]\mu = \frac{r^2 \times \Delta P}{8lv} \\\\\mu = \frac{(2.5\times 10^{-6})^2 \times 2850}{8 \times 2\times 10^{-3} \times 0.0012} \\\\\mu= 9.27 \times 10^{-4} \ Pa.s[/tex]

Thus, the viscosity of the blood flowing in the given capillary is [tex]9.27 \times 10^{-4} \ Pa.s[/tex]

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